
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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Transcribed Image Text:**Problem 1:**
The transistor in the circuit of Figure 6.3.1 has a current gain (β) of 100. Find the values of the four resistors so that the following conditions are met:
- \( V_B = +5 \, V \)
- \( I_1 = 0.1 \, mA \)
- \( I_E = 1 \, mA \)
- \( V_C = +6 \, V \)
**Figure 6.3.1 Explanation:**
This figure depicts a transistor biasing circuit with the following components and parameters:
- **Power Supply:**
- \( V_{CC} = +15 \, V \)
- **Transistor Configuration:**
- The circuit includes a bipolar junction transistor.
- **Resistors:**
- \( R_{B1} \) and \( R_{B2} \) are base bias resistors.
- \( R_C \) is the collector resistor.
- \( R_E \) is the emitter resistor.
- **Currents and Voltages:**
- \( I_1 \) is the current through \( R_{B1} \).
- \( V_B \) is the base voltage.
- \( I_E \) is the emitter current.
- \( V_C \) is the collector voltage.
The goal is to determine the resistor values that satisfy the specified conditions.
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- In the circuit shown all transistors are identical with B = 80. We give R1 such that lo = 4 mA , R2 = 0, and Rc = 4 kQ, then the common mode and difference mode %3D %3D voltage gains and the CMRR are: +10 V Q4 -10 Varrow_forwardDesign the circuit in the figure to obtain I = 1.01 µA, ID = 0.50 mA, V3 = 1.90 V, and Vp = 5.70 V. The NMOS transistor has V; = 0.57 V, K, = 4 mA/V2, and A = 0. Vpp =+10 V Rp VD -o Vs Rs Insert below the value of Rp in kn:arrow_forward100 ΚΩ 18 ΚΩ www +15 V IB + VBE Ic 6.4 ΚΩ + VCE 750 Ω Find the values of IB, IC and VCE for the circuit shown. The transistor is in active mode; assume VBE = 0.7 V and ß = 100arrow_forward
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